Question 210776
Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the Unites States is $110,00. This distribution follows the normal distribution with a standard deviation of $40,000.
if we select a random sample of 50 households what is the standard error of the mean?
Ans: 40,000/sqrt(50)
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what is the expected shape of the distribution of the sample mean
The mean of the sample means is $110,000; standard deviation = 40,000/sqrt(50)
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what is the likelihood of selecting a sample with a mean of at least $112,000
z(112,000) = (112,000-110,000)/[40,000/sqrt(50)] = 0.3536

P(x-bar >=112000) = P(z>=0.3536) = 0.3618...
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what is the likelihood of selecting a sample with a mean of more than $100,000
Find the z value of 100,000
Find the P(z is greater than that value)
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find the likelihood of selecting a sample with a mean of no more than $100,000 but no less than $112,000.
Use the z-value of 100,000 and the z-value of 112,000
Find the probability z is between those values.
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Cheers,
Stan H.